Man on a Railroad Bridge

Date: 09/20/98 at 20:04:26
From: Lee Jackson
Subject: Math
A man is three-eighths of the way across a railroad bridge when he
hears a train behind him. The train's speed is 60mph. He concludes
correctly that he has just enough time to run to either end of the
bridge to prevent being hit by the train. How fast can he run?
I concluded 80mph because I multiplied eight thirds, the multiplicative
inverse of three eighths, by 60mph. If I am not correct, please explain
at length.

Date: 09/21/98 at 16:48:47
From: Doctor Peterson
Subject: Re: Math
Hi, Lee. This is a fascinating problem! I solved it by rather involved
algebra first, but then I saw a much easier way. The work is not much
harder than what you did, but the thought behind it is a lot more
involved. You can't just blindly look for something to divide, but need
to think through what would be proportional to what.
The key is that the man would get to either end just in time, even
though the two ends are different distances from him. How can that be?
Well, the train will get to the end where he started before it gets to
the other end, so he has that much more time to get to the far end.
The time it takes for the train to get from one end to the other will
be the length of the bridge divided by its speed. That has to equal the
difference between the times for the man to get to either end, each of
which is his distance from that end, divided by his speed. This gives
us the equation:
Bridge length 5/8 Bridge length 3/8 Bridge length
------------- = ----------------- - -----------------
Train's speed Man's speed Man's speed
See if you can take it from there.
It's amazing that the answer doesn't depend on the length of the bridge
or how far away the train is. I had to add variables for those when
I used algebra directly.
(By the way, do you know anyone who can run 80 miles per hour?)
- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/